2
$\begingroup$

Let $V=\mathbf{Z}^N$ be a free $\mathbf{Z}$-module of rank $N$. Let $S\subseteq V$ be a fixed finite subset.

Consider the submodule $M:=\langle S\rangle\leq V$ generated by $S$. We know form the general theory that $M$ is a free $\mathbf{Z}$-module of rank $n\leq N$.

Q: Is there a theoritical criterion to determine when it is possible to find a subset $B\subseteq S$ such that $B$ is a $\mathbf{Z}$-basis of $M$ (in otherwords a subset $B\subseteq S$ of cardinality $n$ such that $\langle B\rangle=M$)?

In general, such a subset $B\subseteq S$ needs not to exist. For example take $V=\mathbf{Z}$ and $S=\{2,3\}$. Then $\langle 2,3\rangle=\mathbf{Z}$ has rank one, but $2\mathbf{Z}$ and $3\mathbf{Z}$ are not equal to $\mathbf{Z}$.

If $\#S=r$, then taking the standard basis of $V$, one may associate to $S$ an $r\times N$ matrix. So a possible criterion (here I'm specalutating) could consist (partly) at looking at the gcd of determinants of sufficiently many minors of suitable sizes.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I assume there is a typo in the question; did you intend to ask when you can find a $Z$-basis $B$ for $M$ that is part of a $Z$-basis of $V$? (That is, when is $M$ a direct summand of $V$?) Necessary and sufficient for this to happen is that the GCD of all the $k \times k$ minors equals $1$, where $k$ is the rank of $M$. $\endgroup$
    – David Handelman
    Commented Jul 18, 2014 at 14:16
  • $\begingroup$ no, I really meant what I wrote. $\endgroup$
    – Hugo Chapdelaine
    Commented Jul 18, 2014 at 17:46

1 Answer 1

Reset to default
3
$\begingroup$

The condition is the following. Let $A$ be the matrix whose rows are the elements of $S$. Then there is some ordering $v_1,\dots,v_n$ of some $n$ elements of $S$ such that the for all $1\leq k\leq n$, the gcd of the $k\times k$ minors of the matrix whose rows are $v_1,\dots,v_k$ equals the gcd of the $k\times k$ minors of $A$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, now I think I see how to prove it. The point is that multiplication on the left or the right by an invertible matrix with coefficients is Z preserves the determinant of the $k\times k$ minors. $\endgroup$
    – Hugo Chapdelaine
    Commented Jul 19, 2014 at 1:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .